The pKb of the cyanide ion (CN-) is 4.60. What is the pOH and the pH of a 0.05 M cyanide solution?
CN- + H2O ---> HCN + OH-
Kb = [HCN][OH-]/[CN-]
pKb = 4.60
-logKb = 4.60
Kb = antilog(-4.60) = 2.51*10-5
Kb = [HCN][OH-]/[CN-]
2.51*10-5 = x.x/0.05
x2 = 1.25*10-6M
x = (1.25*10-6)1/2 = 1.12*10-3M = [OH-]
pOH = -logOH- = -log(1.12*10-3) = 2.95
pH*pOH = Kw
pH = Kw/pOH
= 10-14/2.95 = 3.38*10-15
pH = 3.38*10-15
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